You need to evaluate the binomial coefficient, using the next formula, such that:
`((n),(k)) = (n!)/(k!(n-k)!)`
Replacing 100 for n and 2 for k yields:
`((100),(2)) = (100!)/(2!(100-2)!) => ((100),(2)) = (100!)/(2!(98)!) `
Notice that `100! = 98!*99*100 `
`((100),(2)) = (98!*99*100)/(2!(98)!) => ((100),(2)) = (99*100)/(1*2) = 4950`
Hence, evaluating the binomial coefficient yields `((100),(2)) = 4950.`
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